Question

A basketball is shot at 12.0 m/s at a 55.0 degree angle. What is the velocity of the ball 2.16 seconds later?

Answer 1

The ball has two projections of speed: $v_{x}$ and $v_{y}$ .

At the initial moment:

$v_{0x}=v_{0}\times \cos(\alpha) \newline v_{0y}=v_{0}\times \sin(\alpha)$

The x-projection of speed will not change in time.

The y-projection:

$v_{y}=v_{0y}+at=v_{0y}-gt$

Take $g=10$ :

$v_{x}=v_{0x}=12\times\cos(55)\approx6.88\newline v_{y}=12\times\sin(55)-10\times2.16\approx-11.76$

So:

$v=\sqrt{v_{x}^2+v_{y}^2}=\sqrt{47.3344+138.2976}\approx13.63m/s$